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Here is the strangest rule of orbital flight: to catch up you slow down, and to go higher you speed up, after which you end up going slower. Walter Hohmann, a German city engineer, worked out in 1925 the cheapest way to move between two circular orbits. Almost every satellite in geostationary orbit got there using his idea.
Speed up to climb, and you arrive slower than you left.
Look
- Burn 1, forwards. The extra speed does not make the satellite go faster around its circle: it stretches the circle into an ellipse. The low point stays where the burn happened; the far side rises to the target height.
- Coast. Half an orbit with no engine. The satellite climbs and, like a ball thrown up, slows down as it goes.
- Burn 2, forwards again. At the top it is now too slow to stay there and would fall back. A second push makes the orbit circular at the new height.
From 300 km to geostationary orbit the two burns add up to about 3.9 km/s, and the trip takes about 5 hours 17 minutes.
Understand
The speed at any point of an orbit comes from vis-viva: v = √(μ(2/r − 1/a)). A Hohmann transfer ellipse touches both circles, so its semi-major axis is a_t = (r₁ + r₂)/2.
| Radius (km) | Speed (km/s) | |
|---|---|---|
| Circular at 300 km | 6,678 | 7.73 |
| Transfer, low point | 6,678 | 10.15 |
| Transfer, high point | 42,164 | 1.61 |
| Circular, geostationary | 42,164 | 3.07 |
Burn 1 = 10.15 − 7.73 ≈ 2.43 km/s. Burn 2 = 3.07 − 1.61 ≈ 1.47 km/s. The coast is half the transfer ellipse’s period: π√(a_t³/μ) ≈ 5.3 hours.
(Real geostationary launches also have to change the orbit’s tilt from the launch site’s latitude to the equator, which costs more; launching near the equator saves fuel.)
Master
Δv₁ = √(μ/r₁)·(√(2r₂/(r₁+r₂)) − 1), Δv₂ = √(μ/r₂)·(1 − √(2r₁/(r₁+r₂)))
The Hohmann transfer is the minimum two-impulse transfer between coplanar circular orbits, as long as r₂/r₁ < 11.94. Beyond that ratio a bi-elliptic transfer (three burns, going beyond the target first) costs less Δv, at the price of much more time.
Burning at the low point is also the most efficient place to add energy: the Oberth effect says a given Δv changes the orbital energy most where the spacecraft is moving fastest, because Δε ≈ v·Δv.
Probeer het eens
Use vis-viva to check the speed at the top of the transfer: μ = 398,600 km³/s², r = 42,164 km, a = 24,421 km. Then compare it with the circular speed at 42,164 km. The difference is the second burn.
Uitdagingen zijn waar het leren aan vasthoudt.
Een snelle quiz
3 snelle vragen. Kies een antwoord om te zien of je gelijk hebt.
-
To reach a higher circular orbit, the first burn is:
- A Straight up
- B Forwards, speeding up
- C Backwards, slowing down
- D Sideways, out of the orbit plane
Toon het antwoord
B. Forwards, speeding up
-
During the coast between the two burns, the satellite:
- A Speeds up as it climbs
- B Slows down as it climbs
- C Keeps the same speed
- D Stops at the top
Toon het antwoord
B. Slows down as it climbs Climbing away from Earth trades kinetic energy for potential energy, like a ball thrown upwards.
-
Why is the geostationary orbit useful?
- A It is closest to Earth
- B A satellite there takes one day per orbit, so it stays over the same spot
- C There is no gravity there
- D It is the fastest orbit
Toon het antwoord
B. A satellite there takes one day per orbit, so it stays over the same spot
De finishlijn
- Lees de les
- De uitdaging gedaan
- De quiz gedaan
Markeer de les als voltooid om deze op te slaan in je voortgang op dit apparaat.
Les voltooid. Goed gedaan!
Volgende pagina » Launch windows: the road to MarsWoorden in deze les
Bronnen
- W. Hohmann, Die Erreichbarkeit der Himmelskörper (1925), NASA technical translation F-44
- NASA JPL, Basics of Space Flight, Chapter 4: Trajectories
- JPL SSD, Astrodynamic parameters
- R. Bate, D. Mueller, J. White, Fundamentals of Astrodynamics (Dover, 1971), ch. 3
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